HI, It is a brilliant opportunity to take this course. I’m especially looking forward to learning how to utilize R and GitHub efficiently in my research process. I found the course through the Weboodi course catalog that lists mandatory interdisciplinary Ph.D. courses.
Link to the GitHub repository page
# This is a so-called "R chunk" where you can write R code.
date()
## [1] "Tue Nov 17 20:14:52 2020"
The text continues here.
Today is
date()
## [1] "Tue Nov 17 20:14:52 2020"
Let’s start by reading the data in R.
learning2014 <- read.csv("/Users/marttikaila/IODS-project/data/learning2014.csv",row.names = 1)
Lets then provide some descriptive statistics
summary(learning2014)
## Attitude gender Age Stra
## Min. :1.400 Min. :0.0000 Min. :17.00 Min. :1.250
## 1st Qu.:2.600 1st Qu.:0.0000 1st Qu.:21.00 1st Qu.:2.625
## Median :3.200 Median :1.0000 Median :22.00 Median :3.188
## Mean :3.143 Mean :0.6627 Mean :25.51 Mean :3.121
## 3rd Qu.:3.700 3rd Qu.:1.0000 3rd Qu.:27.00 3rd Qu.:3.625
## Max. :5.000 Max. :1.0000 Max. :55.00 Max. :5.000
## Deep Points Surf
## Min. :1.583 Min. : 7.00 Min. :1.583
## 1st Qu.:3.333 1st Qu.:19.00 1st Qu.:2.417
## Median :3.667 Median :23.00 Median :2.833
## Mean :3.680 Mean :22.72 Mean :2.787
## 3rd Qu.:4.083 3rd Qu.:27.75 3rd Qu.:3.167
## Max. :4.917 Max. :33.00 Max. :4.333
Based on the descriptive statistics, the sample contains relatively young people, and most individuals are female. Since I coded the gender variable as a binary variable that takes value one if the person is female, we can conclude based on the mean of gender variable that around 66 % of the sample are women. On average, an individual’s age is about 22.5 years. However, a median person in the sample is 3.5 years younger.
Let’s start the graphical overview by plotting the histograms of our variables.
library(ggplot2)
library(gridExtra)
age <- ggplot(learning2014, aes(x = Age)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
points <- ggplot(learning2014, aes(x = Points)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
surf <- ggplot(learning2014, aes(x = Surf)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
attitude <- ggplot(learning2014, aes(x = Attitude)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
stra <- ggplot(learning2014, aes(x = Stra)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
deep <- ggplot(learning2014, aes(x = Deep)) +
geom_histogram(aes(y = ..density..),
color = "grey30", fill = "darkorange") +
geom_density(alpha = .3, fill = "darkorange")
grid.arrange(age, points, surf, attitude, stra, deep , ncol = 2)
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Based on the histograms, we can state that most of the variables have a sort of nice bell-shaped distribution. Using eyeballing statistics, it seems that only the distributions of age and exam scores variables are a bit more skewed. Age distribution has a long right tail. Also, the distribution of points is leaning more towards higher points.
Let then create some scatterplots that show the raw relationship between the points variable and other variables in the data
library(ggplot2)
library(gridExtra)
ggp1 <- ggplot(learning2014, aes(x=Age, y=Points)) +
geom_point()+
geom_smooth(method=lm, color = "grey30", fill = "darkorange")
ggp2 <- ggplot(learning2014, aes(x=Surf, y=Points)) +
geom_point()+
geom_smooth(method=lm, color = "grey30", fill = "darkorange")
ggp3 <- ggplot(learning2014, aes(x=Attitude, y=Points)) +
geom_point()+
geom_smooth(method=lm, color = "grey30", fill = "darkorange")
ggp4 <- ggplot(learning2014, aes(x=Stra, y=Points)) +
geom_point()+
geom_smooth(method=lm, color = "grey30", fill = "darkorange")
ggp5 <- ggplot(learning2014, aes(x=Deep, y=Points)) +
geom_point()+
geom_smooth(method=lm, color = "grey30", fill = "darkorange")
grid.arrange(ggp1, ggp2, ggp3, ggp4, ggp5, ncol = 2)
## `geom_smooth()` using formula 'y ~ x'
## `geom_smooth()` using formula 'y ~ x'
## `geom_smooth()` using formula 'y ~ x'
## `geom_smooth()` using formula 'y ~ x'
## `geom_smooth()` using formula 'y ~ x'
The figure above shows the relationship between points variables and five other variables in the data. The figure hints that the exam points are positively associated with variables attitude and stra. In other words, individuals with higher attitudes and stra scores tend to do better in the exam. The scatterplots and the lines I fit on the top of the dots suggest that the association between attitude and exam score is more robust than the association between stra and exam scores.
On the contrary, the correlations between exam points and variables age, surf, and deep are either negative or zero. The figure provides strong suggestive evidence that there is no relationship between exam points and deep variables. However, the surf and age variables might be negatively associated with exam points.
I decide to pick the following three variables: attitude, stra, and age. I then fit the linear OLS regression model
linearMod <- lm(Points ~ Attitude + Stra + Surf , data=learning2014)
summary(linearMod)
##
## Call:
## lm(formula = Points ~ Attitude + Stra + Surf, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.0171 3.6837 2.991 0.00322 **
## Attitude 3.3952 0.5741 5.913 1.93e-08 ***
## Stra 0.8531 0.5416 1.575 0.11716
## Surf -0.5861 0.8014 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08
Summary of the regression analysis results shows the variables attitude and Stra are positively associated with exam points while the surf variable has a negative relationship with exam performance. However, the summary also reveals that only the attitude regression coefficient is statistically significant at a 95 percent confidence level, although based on the F-statistic, all the variables are jointly significant. In any case, I only keep the attitude variable and re-run the analysis.
linearMod <- lm(Points ~ Attitude , data=learning2014)
summary(linearMod)
##
## Call:
## lm(formula = Points ~ Attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.6372 1.8303 6.358 1.95e-09 ***
## Attitude 3.5255 0.5674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09
We get relatively similar results with the model that includes just the attitude variables. The regression coefficient of the attitude variable and the fit of the model are smaller, but differences are tiny. There are a couple of ways to interpret the model. First, the regressor’s coefficient tells us how much does the expected exam score increases on average if the attitude score increases by one point. Hence, our model says that if the attitude score increases by 1 point expected exam score should increase by 3.5 points. Also, using the intercept term and regression coefficient together, we can calculate what the expected exam score conditional on an individual’s attitude score.
Second, R2 or the fit of the model tells us how large share in the variance of the exam points is explained by our model. In my case, R squared is around 0.185, meaning that around 18 percent of the variance in the exam scores can be explained by the model i.e attitude scores. Hence, considering that we are doing social science where models have relatively low fit compared to, for example, natural sciences, the attitude seems to do well at predicting exam scores.
The model’s coefficient presented above can be interpreted as the best linear unbiased estimate, and the statistical tests that I examined are valid if the following conditions hold.
We can evaluate some of these assumptions by looking at some diagnostic plots that were asked for. Let us plot these figures
plot(lm(Points ~Attitude , data=learning2014))
The first figure plots relationship between the residuals of the model and fitted values. We get fitted value by using our model to predict the value of the outcome variable, whereas residual is the difference between the fitted value and observed value. The residuals vs fitted value figure can be used to evaluate whether the linearity or constant variance assumptions hold. If the assumptions hold, the dots in the figure should be evenly distributed around the horizontal line. However, the figure seems to the point that, on average, dots get closer to the line at the larger fitted values.
In the previous part of the exercise, we were using t-statistics and F-statistics to evaluate if our estimates were statistically significant. In a small sample, this is a valid approach if the error term of the model is normally distributed. In the second figure above, we are evaluating whether the normality assumption holds by comparing theoretical normal distribution to the observed distribution of residuals. If these distributions are entirely similar, all dots should along a 45-degree line that goes through the graph. In our case, some dots in the lowest and highest decile are off the line, indicating that the error term’s distribution is a bit skewed.
Finally, we want to check whether there are any influential outlier observations that might significantly impact estimates. OLS estimator finds an estimate that minimizes the sum of squared residuals of the model. Because residuals are squared, the outliers that are far away from other observations might significantly impact estimates and create bias. In the last figure, we test how influential every single observation is. Based on the figure, it turns out that outliers do not seem to be a big problem.
Today is
date()
## [1] "Tue Nov 17 20:14:59 2020"
Done!
Let’s start by reading the data in R.
joinedData <- read.csv("/Users/marttikaila/IODS-project/data/joinedData.csv",row.names = 1)
Let’s star by providing glimpse to our data
library(dplyr)
##
## Attaching package: 'dplyr'
## The following object is masked from 'package:gridExtra':
##
## combine
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
glimpse(joinedData)
## Rows: 382
## Columns: 35
## $ school <chr> "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP"…
## $ sex <chr> "F", "F", "F", "F", "F", "M", "M", "F", "M", "M", "F", "F"…
## $ age <int> 18, 17, 15, 15, 16, 16, 16, 17, 15, 15, 15, 15, 15, 15, 15…
## $ address <chr> "U", "U", "U", "U", "U", "U", "U", "U", "U", "U", "U", "U"…
## $ famsize <chr> "GT3", "GT3", "LE3", "GT3", "GT3", "LE3", "LE3", "GT3", "L…
## $ Pstatus <chr> "A", "T", "T", "T", "T", "T", "T", "A", "A", "T", "T", "T"…
## $ Medu <int> 4, 1, 1, 4, 3, 4, 2, 4, 3, 3, 4, 2, 4, 4, 2, 4, 4, 3, 3, 4…
## $ Fedu <int> 4, 1, 1, 2, 3, 3, 2, 4, 2, 4, 4, 1, 4, 3, 2, 4, 4, 3, 2, 3…
## $ Mjob <chr> "at_home", "at_home", "at_home", "health", "other", "servi…
## $ Fjob <chr> "teacher", "other", "other", "services", "other", "other",…
## $ reason <chr> "course", "course", "other", "home", "home", "reputation",…
## $ nursery <chr> "yes", "no", "yes", "yes", "yes", "yes", "yes", "yes", "ye…
## $ internet <chr> "no", "yes", "yes", "yes", "no", "yes", "yes", "no", "yes"…
## $ guardian <chr> "mother", "father", "mother", "mother", "father", "mother"…
## $ traveltime <int> 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 1…
## $ studytime <int> 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 3, 1, 3, 2, 1, 1…
## $ failures <int> 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0…
## $ schoolsup <chr> "yes", "no", "yes", "no", "no", "no", "no", "yes", "no", "…
## $ famsup <chr> "no", "yes", "no", "yes", "yes", "yes", "no", "yes", "yes"…
## $ paid <chr> "no", "no", "no", "no", "no", "no", "no", "no", "no", "no"…
## $ activities <chr> "no", "no", "no", "yes", "no", "yes", "no", "no", "no", "y…
## $ higher <chr> "yes", "yes", "yes", "yes", "yes", "yes", "yes", "yes", "y…
## $ romantic <chr> "no", "no", "no", "yes", "no", "no", "no", "no", "no", "no…
## $ famrel <int> 4, 5, 4, 3, 4, 5, 4, 4, 4, 5, 3, 5, 4, 5, 4, 4, 3, 5, 5, 3…
## $ freetime <int> 3, 3, 3, 2, 3, 4, 4, 1, 2, 5, 3, 2, 3, 4, 5, 4, 2, 3, 5, 1…
## $ goout <int> 4, 3, 2, 2, 2, 2, 4, 4, 2, 1, 3, 2, 3, 3, 2, 4, 3, 2, 5, 3…
## $ Dalc <int> 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1…
## $ Walc <int> 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1, 2, 2, 1, 4, 3…
## $ health <int> 3, 3, 3, 5, 5, 5, 3, 1, 1, 5, 2, 4, 5, 3, 3, 2, 2, 4, 5, 5…
## $ absences <int> 5, 3, 8, 1, 2, 8, 0, 4, 0, 0, 1, 2, 1, 1, 0, 5, 8, 3, 9, 5…
## $ G1 <int> 2, 7, 10, 14, 8, 14, 12, 8, 16, 13, 12, 10, 13, 11, 14, 16…
## $ G2 <int> 8, 8, 10, 14, 12, 14, 12, 9, 17, 14, 11, 12, 14, 11, 15, 1…
## $ G3 <int> 8, 8, 11, 14, 12, 14, 12, 10, 18, 14, 12, 12, 13, 12, 16, …
## $ alc_use <dbl> 1.0, 1.0, 2.5, 1.0, 1.5, 1.5, 1.0, 1.0, 1.0, 1.0, 1.5, 1.0…
## $ high_use <lgl> FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE, FALSE, FAL…
So we have 382 separate observations and 35 variables measuring different things. The data have roughly two categories of variables. First, there are variables measuring background characteristics that only related to individual. These are, for example, individual’s sex, age, school performance and alcohol consumption. Second, there are variables that are related to individuals parental background. These are, for example, family size, parents’ cohabitation status, and parental education status.
summary(joinedData)
## school sex age address
## Length:382 Length:382 Min. :15.00 Length:382
## Class :character Class :character 1st Qu.:16.00 Class :character
## Mode :character Mode :character Median :17.00 Mode :character
## Mean :16.59
## 3rd Qu.:17.00
## Max. :22.00
## famsize Pstatus Medu Fedu
## Length:382 Length:382 Min. :0.000 Min. :0.000
## Class :character Class :character 1st Qu.:2.000 1st Qu.:2.000
## Mode :character Mode :character Median :3.000 Median :3.000
## Mean :2.806 Mean :2.565
## 3rd Qu.:4.000 3rd Qu.:4.000
## Max. :4.000 Max. :4.000
## Mjob Fjob reason nursery
## Length:382 Length:382 Length:382 Length:382
## Class :character Class :character Class :character Class :character
## Mode :character Mode :character Mode :character Mode :character
##
##
##
## internet guardian traveltime studytime
## Length:382 Length:382 Min. :1.000 Min. :1.000
## Class :character Class :character 1st Qu.:1.000 1st Qu.:1.000
## Mode :character Mode :character Median :1.000 Median :2.000
## Mean :1.448 Mean :2.037
## 3rd Qu.:2.000 3rd Qu.:2.000
## Max. :4.000 Max. :4.000
## failures schoolsup famsup paid
## Min. :0.0000 Length:382 Length:382 Length:382
## 1st Qu.:0.0000 Class :character Class :character Class :character
## Median :0.0000 Mode :character Mode :character Mode :character
## Mean :0.2016
## 3rd Qu.:0.0000
## Max. :3.0000
## activities higher romantic famrel
## Length:382 Length:382 Length:382 Min. :1.000
## Class :character Class :character Class :character 1st Qu.:4.000
## Mode :character Mode :character Mode :character Median :4.000
## Mean :3.937
## 3rd Qu.:5.000
## Max. :5.000
## freetime goout Dalc Walc health
## Min. :1.00 Min. :1.000 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:3.00 1st Qu.:2.000 1st Qu.:1.000 1st Qu.:1.000 1st Qu.:3.000
## Median :3.00 Median :3.000 Median :1.000 Median :2.000 Median :4.000
## Mean :3.22 Mean :3.113 Mean :1.482 Mean :2.296 Mean :3.573
## 3rd Qu.:4.00 3rd Qu.:4.000 3rd Qu.:2.000 3rd Qu.:3.000 3rd Qu.:5.000
## Max. :5.00 Max. :5.000 Max. :5.000 Max. :5.000 Max. :5.000
## absences G1 G2 G3 alc_use
## Min. : 0.0 Min. : 2.00 Min. : 4.00 Min. : 0.00 Min. :1.000
## 1st Qu.: 1.0 1st Qu.:10.00 1st Qu.:10.00 1st Qu.:10.00 1st Qu.:1.000
## Median : 3.0 Median :12.00 Median :12.00 Median :12.00 Median :1.500
## Mean : 4.5 Mean :11.49 Mean :11.47 Mean :11.46 Mean :1.889
## 3rd Qu.: 6.0 3rd Qu.:14.00 3rd Qu.:14.00 3rd Qu.:14.00 3rd Qu.:2.500
## Max. :45.0 Max. :18.00 Max. :18.00 Max. :18.00 Max. :5.000
## high_use
## Mode :logical
## FALSE:268
## TRUE :114
##
##
##
I pick the following four variables:
Let’s create the data that contain only these independent variables and then alcohol related outcomes.
library(tidyr); library(dplyr); library(ggplot2); library(GGally)
analysisAlc <- joinedData %>%
select(sex, age, Fedu, absences, high_use)
analysisAlc$Fedu <- cut(analysisAlc$Fedu, breaks = c(-Inf, 0, 1, 2, 3, Inf), labels= c("None", "primaryLow", "primaryHigh", "secondary", "Tertiary") )
analysisAlc$Fedu <- factor(analysisAlc$Fedu)
summary(analysisAlc)
## sex age Fedu absences
## Length:382 Min. :15.00 None : 2 Min. : 0.0
## Class :character 1st Qu.:16.00 primaryLow : 77 1st Qu.: 1.0
## Mode :character Median :17.00 primaryHigh:105 Median : 3.0
## Mean :16.59 secondary : 99 Mean : 4.5
## 3rd Qu.:17.00 Tertiary : 99 3rd Qu.: 6.0
## Max. :22.00 Max. :45.0
## high_use
## Mode :logical
## FALSE:268
## TRUE :114
##
##
##
library(tidyr); library(dplyr); library(ggplot2); library(GGally)
gather(analysisAlc) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
Based on the histogram, around 1/3 of the sample are high users of alcohol. Otherwise, the sample individuals are relatively young, and their fathers are somewhat educated. Few individuals are 19-21 years old. If these individuals are in the sample because of school retention, any associations we find between age and heavy drinking may be explained by other unobserved problems that are behind the school retention. The distribution of absences is very skewed. On the one hand, there are a lot of individuals who are almost never absent, and on the other hand, quite many individuals whose absence numbers are high.
library(tidyr); library(dplyr); library(ggplot2); library(GGally)
box1 <- ggplot(analysisAlc, aes(x = high_use , y = absences, col = Fedu ))
box1 + geom_boxplot() + ylab("Absences")
Boxplots are showing many descriptive statistics of absences and age variables for high and low alcohol consumption groups. The line, which is in the middle of the box, reports the median value of the variable on the y-axis. The different boxes within the same box represent different sub-groups.
We can maybe see from the first figure that the median number of absences is larger among high alcohol users. This is in line with my fourth assumption. However, the first box figure also shows that fathers education does not seem to be related to
library(tidyr); library(dplyr); library(ggplot2); library(GGally)
box1 <- ggplot(analysisAlc, aes(x = high_use , y = age, col = Fedu ))
box1 + geom_boxplot() + ylab("age")
The second box figure provides some evidence that the median age is higher among high alcohol users. The figure also points out that there is no relationship between age and father’s education. This is in line with my second assumption.
library(tidyr); library(dplyr); library(ggplot2); library(knitr)
analysisAlc %>%
group_by(Fedu, high_use) %>%
summarise(n=n()) %>%
spread(high_use, n) %>%
kable()
| Fedu | FALSE | TRUE |
|---|---|---|
| None | 2 | NA |
| primaryLow | 53 | 24 |
| primaryHigh | 75 | 30 |
| secondary | 72 | 27 |
| Tertiary | 66 | 33 |
analysisAlc %>%
group_by(Fedu) %>%
summarise(mean_high_use=mean(high_use)) %>%
kable()
| Fedu | mean_high_use |
|---|---|
| None | 0.0000000 |
| primaryLow | 0.3116883 |
| primaryHigh | 0.2857143 |
| secondary | 0.2727273 |
| Tertiary | 0.3333333 |
The first table shows the number of high and low alcohol users by the father’s educational background. The second table shows the fraction of high alcohol users by the father’s educational background. Together the tables provide evidence that there is no relationship between high alcohol use and parental education. This is against my hypothesis.
library(tidyr); library(dplyr); library(ggplot2); library(knitr)
analysisAlc %>%
group_by(sex, high_use) %>%
summarise(n=n()) %>%
spread(high_use, n) %>%
kable()
| sex | FALSE | TRUE |
|---|---|---|
| F | 156 | 42 |
| M | 112 | 72 |
analysisAlc %>%
group_by(sex) %>%
summarise(mean_high_use=mean(high_use)) %>%
kable()
| sex | mean_high_use |
|---|---|
| F | 0.2121212 |
| M | 0.3913043 |
Based on tables 3 and 4, it seems that high alcohol use is more common among males, which supports my first assumption.
Let us fit the model and summarize the results
model <- glm(high_use ~ sex + absences + age + Fedu, data = analysisAlc, family = "binomial")
# summarize the model
summary(model)
##
## Call:
## glm(formula = high_use ~ sex + absences + age + Fedu, family = "binomial",
## data = analysisAlc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2790 -0.8479 -0.6231 1.0561 2.1848
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -18.41011 623.86680 -0.030 0.976
## sexM 1.00410 0.24205 4.148 3.35e-05 ***
## absences 0.09335 0.02326 4.014 5.98e-05 ***
## age 0.16813 0.10351 1.624 0.104
## FeduprimaryLow 13.90961 623.86478 0.022 0.982
## FeduprimaryHigh 13.70235 623.86475 0.022 0.982
## Fedusecondary 13.59776 623.86476 0.022 0.983
## FeduTertiary 13.96070 623.86475 0.022 0.982
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 423.95 on 374 degrees of freedom
## AIC: 439.95
##
## Number of Fisher Scoring iterations: 13
coef(model)
## (Intercept) sexM absences age FeduprimaryLow
## -18.41010883 1.00410227 0.09335241 0.16813210 13.90961364
## FeduprimaryHigh Fedusecondary FeduTertiary
## 13.70235172 13.59776078 13.96070193
We find that all the coefficients are positive, meaning that if an individual is a male, age increases, or the number of absences increases, the probability of high alcohol consumption increases. However, not that the effect of age is not significant. Also, it isn’t easy to interpret what is the impact of a father’s education. However, we can at least conclude that it is not statistically significant.
In order to talk say something more than what is the direction of efffect, let us derive odds ratios for the variables.
# Let's get the odds ratios an the confidence intervals
OR <- coef(model) %>% exp
CI <- confint(model) %>% exp
## Waiting for profiling to be done...
cbind(OR, CI)
## OR 2.5 % 97.5 %
## (Intercept) 1.010628e-08 NA 1.713949e+33
## sexM 2.729456e+00 1.708562e+00 4.421144e+00
## absences 1.097849e+00 1.051069e+00 1.151547e+00
## age 1.183093e+00 9.669145e-01 1.452162e+00
## FeduprimaryLow 1.098673e+06 3.612856e-37 NA
## FeduprimaryHigh 8.930088e+05 2.829941e-37 NA
## Fedusecondary 8.043267e+05 2.570223e-37 NA
## FeduTertiary 1.156261e+06 3.630630e-37 NA
Based on the Confidence intervals age variable is not statistically significant. Also, it seems that the father’s educational background variable is problematic since there is no variation in our outcome variable in some of the education categories. Hence, I drop these two variables and fit the model again.
# Let's get the odds ratios an the confidence intervals
model2 <- glm(high_use ~ sex + absences, data = analysisAlc, family = "binomial")
OR <- coef(model2) %>% exp
CI <- confint(model2) %>% exp
## Waiting for profiling to be done...
cbind(OR, CI)
## OR 2.5 % 97.5 %
## (Intercept) 0.159445 0.1012577 0.2427684
## sexM 2.658116 1.6710354 4.2863129
## absences 1.101409 1.0549317 1.1548057
Now we find that that odds ratio for the male indicator is statistically significant and takes a value 2.6. Interpretation of this is that if a person is a male, the odds of high alcohol consumption is 2.6 times greater when absences are held constant.
The interpretation of the absences’ odds ratio coefficient is a bit different because it is a continuous measure. The odds ratio takes value 1.1 and means that if absences increase by one, the odds of high alcohol usage increases by a factor of 1.1. This implies that if absences increase by n times, the odd of high alcohol usage increases by 1.1^n
Hence, the statistical model seems to confirm two of my hypothesis and reject the other two. However, it is essential to keep in mind that all relationships found above are associations and not causal.
library(dplyr); library(ggplot2); library(visreg); library(caret); library(sjlabelled)
## Loading required package: lattice
##
## Attaching package: 'sjlabelled'
## The following object is masked from 'package:dplyr':
##
## as_label
model2 <- glm(high_use ~ sex + absences, data = analysisAlc, family = "binomial")
# Create propensity score
probs <- predict(model2, type = "response")
# Add propensity score back to the data
analysisAlc<- mutate(analysisAlc, probability = probs)
# Use propebsity scores to predict high_use
analysisAlc <- mutate(analysisAlc, prediction = probability > 0.5)
# Lets summarize and tabulate the high_use variable and our prediction.
table(high_use = analysisAlc$high_use, prediction = analysisAlc$prediction)
## prediction
## high_use FALSE TRUE
## FALSE 258 10
## TRUE 88 26
table(high_use = analysisAlc$high_use, prediction = analysisAlc$prediction) %>%
prop.table() %>% addmargins
## prediction
## high_use FALSE TRUE Sum
## FALSE 0.67539267 0.02617801 0.70157068
## TRUE 0.23036649 0.06806283 0.29842932
## Sum 0.90575916 0.09424084 1.00000000
# initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(analysisAlc, aes(x = probability, y = high_use , col = prediction))
g + geom_point()
Based on the figure and confusion matrix, it seems that our model produces quite many false-negative predictions meaning that our model says that the person is not a high user of alcohol when the person actually is a high user.
Let me also draw these cool figures that show how the marginal probability of high user evolves as a function of absences and how this relationship varies between gender.
# define the geom as points and draw the plot
visreg(model2 , "absences",
gg = TRUE,
scale="response") +
labs(y = "Prob(High Use)",
x = "absences",
title = "Relationship of absences and High Use",
subtitle = "controlling for sex"
)
visreg(model2 , "absences",
gg = TRUE,
by = "sex",
scale="response") +
labs(y = "Prob(High Use)",
x = "absences",
title = "Relationship of absences and High Use",
subtitle = "controlling for sex"
)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
loss_func(class = analysisAlc$high_use, prob = analysisAlc$probability)
## [1] 0.2565445
It seems that 25 percent of our predictions are wrong. Let’s compare this to a strategy in which we say that every third person is a high alcohol user.
loss_func(class = analysisAlc$high_use, prob = 0.333)
## [1] 0.2984293
With this model, around 1/3 of our predictions are wrong. Hence, the proper statistical model does just a little bit better than the random guessing strategy.
Today is
date()
## [1] "Tue Nov 17 20:15:06 2020"
done!
Let us load the data and then summarize and provide a glimpse into the data.
library(dplyr); library (MASS)
glimpse(Boston)
## Rows: 506
## Columns: 14
## $ crim <dbl> 0.00632, 0.02731, 0.02729, 0.03237, 0.06905, 0.02985, 0.08829…
## $ zn <dbl> 18.0, 0.0, 0.0, 0.0, 0.0, 0.0, 12.5, 12.5, 12.5, 12.5, 12.5, …
## $ indus <dbl> 2.31, 7.07, 7.07, 2.18, 2.18, 2.18, 7.87, 7.87, 7.87, 7.87, 7…
## $ chas <int> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0…
## $ nox <dbl> 0.538, 0.469, 0.469, 0.458, 0.458, 0.458, 0.524, 0.524, 0.524…
## $ rm <dbl> 6.575, 6.421, 7.185, 6.998, 7.147, 6.430, 6.012, 6.172, 5.631…
## $ age <dbl> 65.2, 78.9, 61.1, 45.8, 54.2, 58.7, 66.6, 96.1, 100.0, 85.9, …
## $ dis <dbl> 4.0900, 4.9671, 4.9671, 6.0622, 6.0622, 6.0622, 5.5605, 5.950…
## $ rad <int> 1, 2, 2, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4…
## $ tax <dbl> 296, 242, 242, 222, 222, 222, 311, 311, 311, 311, 311, 311, 3…
## $ ptratio <dbl> 15.3, 17.8, 17.8, 18.7, 18.7, 18.7, 15.2, 15.2, 15.2, 15.2, 1…
## $ black <dbl> 396.90, 396.90, 392.83, 394.63, 396.90, 394.12, 395.60, 396.9…
## $ lstat <dbl> 4.98, 9.14, 4.03, 2.94, 5.33, 5.21, 12.43, 19.15, 29.93, 17.1…
## $ medv <dbl> 24.0, 21.6, 34.7, 33.4, 36.2, 28.7, 22.9, 27.1, 16.5, 18.9, 1…
summary(Boston)
## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00
The data contain 506 observations, which each apparently represent different census tracts in Boston. The data have 14 variables that describe various characteristics of these census tracts like crime rate, tax rate, and a median value of homes in the census tract.
I start by drawing some scatter plots.
library(tidyr); library(dplyr); library(ggplot2); library(GGally)
gather(Boston) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
Since the data contain so many variables, most of the histograms are relatively small. However, we can at least say that there seems to be quite a bit of variation in age, nox and medv variables. ### 3.1 Scatter and correlations plots
library("ggplot2"); library("GGally");library(corrplot)
pairs(Boston)
cMatrix = cor(Boston) %>% round(digits=2)
corrplot(cMatrix, method = "ellipse", type= "upper")
Since the correlation plot with scatter plots is messy, I will concentrate just on the latter figure, which shows only the correlations between variables. Based on the figure, some variables like lower statuts of the population (lstat) and median value of homes (medv) are very negatively correlated, some variables like proportion of non-retail business (indus) and emissons (nox) are very positively correlated, whereas between some variables like there do not exist any relationship.
library("ggplot2"); library("GGally");library(corrplot)
bostonStand <-scale(Boston)
bostonStand <- as.data.frame(bostonStand)
summary(bostonStand)
## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865
We can see that after the standardization procedure, each variable is centered around zero. (mean is zero). Let us then create a new categorical variable using the existing crime variable.
library("ggplot2"); library("GGally");library(corrplot); library(dplyr)
bins <- quantile(bostonStand$crim)
crime <-cut(bostonStand$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high") )
bostonStand <- dplyr::select(bostonStand , -crim)
summary(bostonStand)
## zn indus chas nox
## Min. :-0.48724 Min. :-1.5563 Min. :-0.2723 Min. :-1.4644
## 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723 1st Qu.:-0.9121
## Median :-0.48724 Median :-0.2109 Median :-0.2723 Median :-0.1441
## Mean : 0.00000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723 3rd Qu.: 0.5981
## Max. : 3.80047 Max. : 2.4202 Max. : 3.6648 Max. : 2.7296
## rm age dis rad
## Min. :-3.8764 Min. :-2.3331 Min. :-1.2658 Min. :-0.9819
## 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049 1st Qu.:-0.6373
## Median :-0.1084 Median : 0.3171 Median :-0.2790 Median :-0.5225
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617 3rd Qu.: 1.6596
## Max. : 3.5515 Max. : 1.1164 Max. : 3.9566 Max. : 1.6596
## tax ptratio black lstat
## Min. :-1.3127 Min. :-2.7047 Min. :-3.9033 Min. :-1.5296
## 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049 1st Qu.:-0.7986
## Median :-0.4642 Median : 0.2746 Median : 0.3808 Median :-0.1811
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332 3rd Qu.: 0.6024
## Max. : 1.7964 Max. : 1.6372 Max. : 0.4406 Max. : 3.5453
## medv
## Min. :-1.9063
## 1st Qu.:-0.5989
## Median :-0.1449
## Mean : 0.0000
## 3rd Qu.: 0.2683
## Max. : 2.9865
bostonStand$crime <- crime
summary(bostonStand)
## zn indus chas nox
## Min. :-0.48724 Min. :-1.5563 Min. :-0.2723 Min. :-1.4644
## 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723 1st Qu.:-0.9121
## Median :-0.48724 Median :-0.2109 Median :-0.2723 Median :-0.1441
## Mean : 0.00000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723 3rd Qu.: 0.5981
## Max. : 3.80047 Max. : 2.4202 Max. : 3.6648 Max. : 2.7296
## rm age dis rad
## Min. :-3.8764 Min. :-2.3331 Min. :-1.2658 Min. :-0.9819
## 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049 1st Qu.:-0.6373
## Median :-0.1084 Median : 0.3171 Median :-0.2790 Median :-0.5225
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617 3rd Qu.: 1.6596
## Max. : 3.5515 Max. : 1.1164 Max. : 3.9566 Max. : 1.6596
## tax ptratio black lstat
## Min. :-1.3127 Min. :-2.7047 Min. :-3.9033 Min. :-1.5296
## 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049 1st Qu.:-0.7986
## Median :-0.4642 Median : 0.2746 Median : 0.3808 Median :-0.1811
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332 3rd Qu.: 0.6024
## Max. : 1.7964 Max. : 1.6372 Max. : 0.4406 Max. : 3.5453
## medv crime
## Min. :-1.9063 low :127
## 1st Qu.:-0.5989 med_low :126
## Median :-0.1449 med_high:126
## Mean : 0.0000 high :127
## 3rd Qu.: 0.2683
## Max. : 2.9865
Let us create training and test sets and then have a glimpse at both of them.
library("ggplot2");library(corrplot); library(dplyr)
nObs <- nrow(bostonStand)
indTrain <- sample(nObs , size = nObs * 0.8)
# Training set
trainBoston <- bostonStand[indTrain,]
# Test set
testBoston <- bostonStand[-indTrain,]
glimpse(trainBoston)
## Rows: 404
## Columns: 14
## $ zn <dbl> -0.48724019, -0.48724019, 3.58608778, -0.48724019, -0.4872401…
## $ indus <dbl> 0.1156240, 1.2307270, -1.2327031, 1.0149946, 1.2307270, 1.014…
## $ chas <dbl> -0.2723291, -0.2723291, -0.2723291, -0.2723291, -0.2723291, -…
## $ nox <dbl> 0.1579678, 2.7296452, -1.1960462, 0.6584956, 2.7296452, 1.253…
## $ rm <dbl> 0.98398626, -0.22008342, 2.23217669, -1.87105367, -1.93367668…
## $ age <dbl> 0.796660955, 1.116389695, -1.256708066, 1.116389695, 0.963630…
## $ dis <dbl> -0.7729187, -1.1283332, 0.6282713, -1.1694595, -1.1085299, -0…
## $ rad <dbl> -0.9818712, -0.5224844, -0.6373311, 1.6596029, -0.5224844, 1.…
## $ tax <dbl> -0.80241764, -0.03107419, -1.09315478, 1.52941294, -0.0310741…
## $ ptratio <dbl> 1.17530274, -1.73470120, -1.73470120, 0.80577843, -1.73470120…
## $ black <dbl> 0.4406159, -2.0128627, 0.3954874, 0.2064297, 0.3837671, 0.237…
## $ lstat <dbl> -0.98207574, 2.12110437, -1.23834016, -1.31535952, 2.36336527…
## $ medv <dbl> 0.148654801, -0.949516961, 2.823409785, 2.986504601, -0.86253…
## $ crime <fct> low, med_high, low, high, med_high, high, med_high, med_high,…
glimpse(testBoston)
## Rows: 102
## Columns: 14
## $ zn <dbl> -0.4872402, -0.4872402, -0.4872402, -0.4872402, -0.4872402, 2…
## $ indus <dbl> -0.4368257, -0.4368257, -0.4368257, -0.4368257, -0.7545936, -…
## $ chas <dbl> -0.2723291, -0.2723291, -0.2723291, -0.2723291, -0.2723291, -…
## $ nox <dbl> -0.14407485, -0.14407485, -0.14407485, -0.14407485, -0.480636…
## $ rm <dbl> -0.47769171, -0.51327297, -0.97582929, -0.26847393, -0.631402…
## $ age <dbl> -0.24068118, 0.90678974, 0.60837625, 1.00626091, -0.25489135,…
## $ dis <dbl> 0.43332522, 0.28710377, 0.31322322, -0.01673672, -0.19810072,…
## $ rad <dbl> -0.6373311, -0.6373311, -0.6373311, -0.6373311, -0.5224844, -…
## $ tax <dbl> -0.60068166, -0.60068166, -0.60068166, -0.60068166, -0.766817…
## $ ptratio <dbl> 1.17530274, 1.17530274, 1.17530274, 1.17530274, 0.34387304, -…
## $ black <dbl> 0.440615895, 0.412465352, -0.583319029, -1.186967442, 0.22877…
## $ lstat <dbl> -0.61518350, 0.51069953, 0.54010692, 1.07644175, -0.17407261,…
## $ medv <dbl> -0.23189977, -0.75380318, -0.93864397, -0.98213592, -0.275391…
## $ crime <fct> med_high, med_high, med_high, med_high, med_low, low, med_low…
library("ggplot2");library(corrplot); library(dplyr)
lda.fit <- lda( crime ~ . , data = trainBoston )
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(trainBoston$crime)
plot(lda.fit , dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Let us first save the initial values of the crime variable of test data into a separate vector.
library("ggplot2");library(corrplot); library(dplyr)
# save the initial values of crime vectors
orginalCrime <- testBoston$crime
testBoston <- dplyr::select(testBoston, -crime)
Then let us predict the values of the crime variable in the test data using the model we found using the training data. Finally, we compare the predicted and actual values of the crime variable.
library("ggplot2");library(corrplot); library(dplyr)
lda.pred <- predict(lda.fit, newdata = testBoston)
table(correct = orginalCrime, lda.pred$class )
##
## correct low med_low med_high high
## low 15 5 0 0
## med_low 3 16 8 0
## med_high 0 5 18 1
## high 0 0 0 31
The diagonal line from the left top corner to the bottom right corner shows the number of correct predictions. Based on the table, our model performs ok. The model does very well at predicting extreme values (low and high crimes) while there seem to be more wrong guesses when predicting values are closer to the median.
First we re-load the data and standardize the variables. Then I calculate euclidean distances between variables.
library(dplyr); library (MASS)
boston<-scale(Boston)
boston <- as.data.frame(boston)
distE <- dist(boston)
summary(distE )
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970
I first perform k-mean clustering with 7 clusters. Note that I pick the number of clusters just randomly.
library(dplyr); library (MASS)
kM <- kmeans(Boston, centers = 7)
pairs(boston[1:5], col = kM$cluster )
pairs(boston[6:10], col = kM$cluster )
pairs(boston[11:14], col = kM$cluster )
Let us then be a bit formal and find the optimal number of clusters by calculating the total of within-cluster sum of squares (WCSS).
library(dplyr); library (MASS); library("ggplot2");
set.seed(123)
k_max <- 10
twcss <- sapply(1:k_max, function(k){kmeans(boston, k)$tot.withinss})
qplot(x = 1:k_max, y = twcss, geom = 'line')
Based on the figure, it seems that the WCSS drops heavily when the number of clusters is 2. Let us perform the k-mean clustering using two clusters and plot the results
library(dplyr); library (MASS)
kM <- kmeans(Boston, centers = 2)
pairs(boston[1:5], col = kM$cluster )
pairs(boston[6:10], col = kM$cluster )
pairs(boston[11:14], col = kM$cluster )